Question

Find the Maclaurin series for f(x) = sin(x^4).
Help...I'm completely stuck and confused :/

Answer 1

Are we allowed to use the general form for the Maclaurin series and just plug stuff in?\[\Large\rm \sum_{0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n\]

Answer 2

Yes. :)

Answer 3

So I guess calculating derivatives is going to be a little bit of a pain due to the chain rule :(
Hmm.. I wonder if we can just take the sin(x) series and plug x^4 into it or something....
seems like it should work >.<

Answer 4

Yeah..Well I have this so far...
f(x) = sin(x^2);
f'(x) = 2*cos(x^2)*x
f''(x) = -4*sin(x^2)*x^2+2*cos(x^2)
f'''(x) = -8*cos(x^2)*x^3-12*sin(x^2)*x
f^(4)(x) = 16*sin(x^2)*x^4-48*cos(x^2)*x^2-12*sin(x^2)

Answer 5

I think that's right if I did the derivatives correctly.

Answer 6

Are you allowed to cheat and use the series for sin(x)? :O
It works out REALLY nice and simple hehe.

Answer 7

Wait why is it sin(x^2)?
I thought it was sin(x^4), was that a typo?

Answer 8

Ugh oops -_- Didn't think and put 2...Wow...xD

Answer 9

heh

Answer 10

Gotta start over.

Answer 11

I don't know how I would set it up though...Once I get it

Answer 12

If you insist on doing it the long way,
I dunno lemme crunch out some derivatives also and see what we come up with.

Answer 13

Little tip:
if you're writing this all down on paper,
use shorthand for sine and cosine, otherwise it gets really long and messy.

Like I would write f(x)=Sx^4 or something

Answer 14

Since it's a Maclaurin series, we'll need to evaluate each derivative at x=0.
The problem you'll find is.......

The x's don't start to "disappear" from in front of the trig functions until the 5th derivative or so.

So you'll need to calculate to the 7th derivative or something like that...
because the first 4 or so will all be zero :(

Answer 15

Can we just do the shortcut way? >.<

Answer 16

Whats the shortcut way? :)

Answer 17

We can calculate the Maclaurin series for sin(x),
(Taking a few derivatives and such if we need to)

Eventually we'll end up with:\[\Large\rm \sin(x)=x+\frac{x^3}{3!}-\frac{x^5}{5!}+...\]And from there we just plug in x^4 for all the x's.

Answer 18

Really? We can do that? :O

Answer 19

Thats magical xD Hahah I never thought that was like "allowed" xP

Answer 20

I wasn't sure either >.<
I had to look it up.

Seems to be legit though.

Answer 21

Haha this will be fun. xD

Answer 22

Well Im studying for a quiz and this was an example problem. It said to use the series I know.

Answer 23

Ah very cool.

Answer 24

What class?

Answer 25

Physics class, doing some calculus problems to test us.

Answer 26

Like a refresher

Answer 27

Oh ok :O
I dunno what field of study you're going into,
I'm a math major myself.

But if you're going to take any advanced math classes...
make sure you have the power series for sin(x), cos(x) and even e^x TATTOOED into your brain. They show up everywhere.

Answer 28

Thanks haha..Yeah I'm going into computer software and mechanical engineering. I will make sure to get them all down! So..The Maclaurin series expansion for sin(x) is

\sin(x) =\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}x^{2k+1} = x -\frac{x^3}{3!}+\frac{x^5}{5!} - \ldots

and the formula is valid for all real values of x.

Answer 29

Ugh I cant at all use that dumb equation editor xD

Answer 30

lol >.<

Answer 31

So what happens if we replace all the x's with x^4?

Answer 32

I mean...Sinx can be thought of as sinx^1...So I guess it makes sense to get rid of the trouble and think of it like that.

Answer 33

Here is one of the series,\[\Large\rm \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n}}{(2n)!}\]
Here is the other trig function,\[\Large\rm \sum_{n=0}^{\infty}\frac{(-1)^n x^{2n+1}}{(2n+1)!}\]

Here is how I'm able to remember which is which...
Sin(x) is an `odd function`.
Cos(x) is an `even function`.

Can you tell which power series matches up with which using that info? ;)

Answer 34

Sin(x) does :P

Answer 35

I mean...Its not rocket science. sin x turns into cosx after taking the derivative.

Answer 36

Maybe I made that confusing >.<
I posted the series for sin(x) and cos(x) above.
I was trying to see if you could tell which is which :P

Answer 37

Ohhh xD Hahahah Yeah...I miss understood

Answer 38

Sin(x) is odd...Cos(x) is even.

Answer 39

It is known that \(\sin(x) = x - \dfrac{x^3}{3!} + \dfrac{x^5}{5!} - \dfrac{x^7}{7!}+\ldots\) So you can easily replace \(x\) to \(x^4\) and you will get \(\sin(x^4) = (x^4) - \dfrac{(x^4)^3}{3!} + \dfrac{(x^4)^5}{5!} - \ldots = x^4 - \dfrac{x^{12}}{3!} + \dfrac{x^{20}}{5!}-\ldots\).


Hence \[\sin(\color{red}{x}) = \sum_{k=0}^\infty \dfrac{\color{red}{x}^{2k+1}}{2k+1} \\ \sin(\color{blue}{x^4}) = \sum_{k=0}^\infty \dfrac{\color{blue}{x^4}^{2k+1}}{2k+1} = \boxed{\sum_{k=0}^\infty \dfrac{x^{8k+4}}{2k+1}}\]

Answer 40

Oops forgot factorials... haha

Answer 41

and (-1)^k... sorry for careless mistakes... but is it clear, right?

Answer 42

Where does the (-1)^k come in? And that does make a lot of sense! THANK YOU!

Answer 43

Well, (-1)^k have to come in because sign changes from term to term

Answer 44

\(\displaystyle \sin x = \sum_{k=0}^\infty (-1)^k\dfrac{x^{2k+1}}{(2k+1)!}\)

Answer 45

So its the same thing when we write sin(x^4)? It will have the answer you wrote but with (-1)^k

Answer 46

Of course. and (2k+1)! in denominator instead of just 2k+1 :)